Problem: Let $f(x)=x^{-1}$. $f'(x)=$
Explanation: The derivative of $f$ can be found using the power rule : $\dfrac{d}{dx}(x^n)=n\cdot x^{n-1}$ (Remember that this applies even when $n$ is a negative number.) $\begin{aligned} &\phantom{=}f'(x) \\\\ &=\dfrac{d}{dx}\left(x^{-1}\right) \\\\ &=-1x^{-1-1} \gray{\text{The power rule}} \\\\ &=-x^{-2} \end{aligned}$ In conclusion, we found that $f'(x)=-x^{-2}$. This can also be written as $-\dfrac{1}{x^2}$ (all equivalent forms are accepted).